For $n\geq 1$, let $f(n)$ denote the "integer part"
(largest integer below) $n^{\frac{3}{2}}$, and let
$g(n)=f(n+2)-2f(n+1)+f(n)$.
Question : Is it true that $g(n)$ is always in $\lbrace -1,0,1\rbrace$
(excepting the initial value $g(1)=2$) ? I checked this up to
$n=100000$.

It is not too hard to check that if $t$ is large enough compared to $r$ (say
$t\geq \frac{3r^2+1}{4}$),
$f(t^2+r)$ is exactly $t^3+\frac{3rt}{2}$ (or $t^3+\frac{3rt-1}{2}$ if $t$
and $r$ are both odd ) and similarly
$f(t^2-r)$ is exactly $t^3-\frac{3rt}{2}$ (or $t^3-\frac{3rt+1}{2}$ if $t$
and $r$ are both odd ). So we already know that the answer is "yes" for
most of the numbers.

What have you worked out about the analogous non-discretized question? That is, if you put $h(x) = (x+2)^{3/2} - 2(x+1)^{3/2} + x^{3/2}$, can you show that $h$ is bounded by small constants? This would be my first attempt to get a handle on the problem; if it works, then you can examine more closely what happens upon taking "floors". (I don't think the problem has anything really to do with number theory per se.)
–
Yemon ChoiMar 15 '10 at 6:37

@ Yemon : Bjorn Ponnen's method (explained in his answer to my previous question) yields the bound $|h(x)|\leq \frac{c}{\sqrt{x}}$, but it is not clear at all how to deduce a result on the floors from this, apart from the easy (and insufficient) bound $|g(n)| \leq 4+ \frac{c}{\sqrt{n}}$. I added the "number theory" tag beacuse I encountered this problem when working on Hall's conjecture on Mordell's equation
–
Ewan DelanoyMar 15 '10 at 7:36

1

It's still insufficient, but I think you can easily do better than 4. If [x] denotes the floor of x, then (x-1)-2y+(z-1) < [x]-2[y]+[z] < x -2(y-1) + z, which shows that |g(n)-h(n)|<2.
–
Jonas MeyerMar 15 '10 at 7:45

5

One difficulty is that it seems that this would be false over the reals, e.g., for n=20.54 or n=112.675.
–
Douglas ZareMar 15 '10 at 8:01

2

I don't think the range of t and r you gave in the question cover most of the cases. In fact, the asymptotic density of that set appears to be zero.
–
S. Carnahan♦Mar 15 '10 at 9:01

2 Answers
2

Here is a proof that $|g(n)|\le 1$ for all but finitely many $n$. You can extract an explicit bound for $n$ from the argument and check the smaller values by hand.

If $f(n)=n^{3/2}$ without the floor, then $g(n)\sim \frac{3}{4\sqrt n}$, so it is positive and tends to 0. When you replace $n^{3/2}$ by its floor, $g(n)$ changes by at most 2, hence the only chance for failure is to have $g(n)=2$ when the fractional parts of $n^{3/2}$ and $(n+2)^{3/2}$ are very small and the fractional part of $(n+1)^{3/2}$ is very close to 1 (the difference is less than $\frac{const}{\sqrt{n}}$).

Let $a,b,c$ denote the nearest integers to $n^{3/2}$, $(n+1)^{3/2}$ and $(n+2)^{3/2}$. Then $c-2b+a=0$ because it is an integer very close to $(n+2)^{3/2}-2(n+1)^{3/2}+n^{3/2}$. Denote $m=c-b=b-a$. Then $(n+1)^{3/2}-n^{3/2}<m$ and $(n+2)^{3/2}-(n+1)^{3/2}>m$. Observe that
$$
\frac{3}{2}\sqrt{n}<(n+1)^{3/2} - n^{3/2} < \frac{3}{2}\sqrt{n+1}
$$
(the bounds are just the bounds for the derivative of $x^{3/2}$ on $[n,n+1]$. Therefore
$$
\frac{3}{2}\sqrt{n} < m < \frac{3}{2}\sqrt{n+2}
$$
or, equivalently,
$$
n < \frac49 m^2 < n+2.
$$
If $m$ is a multiple of 3, this inequality implies that $n+1=\frac49 m^2=(\frac23m)^2$, then $(n+1)^{3/2}=(\frac23m)^3$ is an integer and not slightly smaller than an integer as it should be. If $m$ is not divisible by 3, then
$$
n+1 = \frac49 m^2 + r
$$
where $r$ is a fraction with denominator 9 and $|r|<1$. From Taylor expansion
$$
f(x+r) = f(x) +r f'(x) +\frac12 r^2 f''(x+r_1), \ \ 0<r_1<r,
$$
for $f(x)=x^{3/2}$, we have
$$
(n+1)^{3/2} = (\frac49 m^2 + r)^{3/2} = \frac8{27}m^3 + mr + \delta
$$
where $0<\delta<\frac1{4m}$.
This cannot be close to an integer because it is close (from above) to a fraction with denominator 27.

It's not too hard to put a bound on the size of second differences (since without the truncation, they are bounded above by a constant times $n^{-1/2}$), but getting the bound down to one seems delicate. It looks like it can be done with mindless brute force, though. I won't write all of the cases, but here is a start. Write $n = t^2 + r$, for integers $t,r$ satisfying $|r| \leq t$. The binomial theorem implies $n^{3/2} = t^3 + (3/2)tr + (3/8)r^2/t - (1/16)r^3/t^3 + (3/128)r^4/t^5 - \dots$.
I'll look at the case where $t$ is even. Then

I've already looked into this way, and it seems that the number of cases and subcases grows hopelessly ; also, the initial "modulo 2" phenomenon (difference between t odd and t even) branches into more complicated phenomena with larger moduli.
–
Ewan DelanoyMar 15 '10 at 8:05